   Chapter 11, Problem 50RE Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

In Exercises 47-52, find the equation of the tangent line to the graph of the given equation at the specified point. y = x 1 + e x ; x = 0

To determine

To calculate: The equation of the tangent to the graph of the equation y=x1+ex at x=0.

Explanation

Given Information:

The provided equation is y=x1+ex and x=0.

Formula used:

Slope of tangent to the graph of f(x) at point (x1,y1) is given by dydx|(x1,y1).

Product rule of derivative of differentiable functions, f(x) and g(x) is

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x).

The derivative of e raised to x is ddxex=ex.

Equation of line is y=mx+b where m is the slope and b=y1mx1 when line passes through (x1,y1).

Calculation:

Consider the equation, y=x1+ex

Find the slope of tangent line to the equation y=x1+ex by determining the derivative.

Then, take ddx of both sides of the equation,

ddx(y)=ddx(x1+ex)dydx=ddx(x1+ex)

Apply the quotient rule of derivative,

dydx=ddx(x)(1+ex)xddx(1+ex)(1+ex)2=1(1+ex)x(ddx(1)+ddxex)(1+ex)2

Simplify the derivative,

dydx=(1+ex)x(0+ex)(1+ex)2=1+exxex(1+ex)2

So, the slope of the tangent to the graph of equation y=x1+ex is 1+exxex(1+ex)2

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